Let (X,ℰ) and (Z,ℱ) be measurable spaces, with ℰ and ℱ the underlying σ-algebras [1]. Consider the mapping

(1) h: (X,ℰ) ↦ (Z,ℱ).

With μX denoting the measure in (X,ℰ), under the mapping h for the measure μZ in (Z,ℱ) one has [1] (p. 39):

(2) μZ = μX ∘ h-1,

where, with f and g denoting mappings, f ∘ g denotes the composite map.

For

(3) h = f ∘ g

one has

(4) h-1 = g-1 ∘ f -1.

Now for (X,ℰ,μX) and (Z,ℱ,μZ) being probability spaces, under the assumption of absolute continuity [1], for the probability distribution functions φX(x) and φZ(z) one has:

(5a) φX(x) = dμX(x)/dx,

(5b) φZ(z) = dμZ(z)/dz.

Following Eq. (2) and in the light of Eqs. (4) and (5), taking into account that (below h'(x) ≔ dh(x)/dx ≡ f '(g(x)) g'(x))

(6) dh-1(z)/dz = 1/h'(h-1(z)),

one obtains:

(7) φZ(z) = φX(h-1(z))/h'(h-1(z)) ≡ φX(g-1(f -1(z))/[h'(g-1(f -1(z))].

Given f and g, and assuming φX(x) to be, say, a Gaussian distribution function, one can immediately determine the distribution function φZ(z).

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[1] A Klenke, Probability Theory: A Comprehensive Course, 2nd edition (Springer, Berlin, 2014).

I am very grateful to you to answer this question.

Does the superscript symbol of “-1” in your high-quality answer denote an inverse mapping or inverse function? I guess so.

To determine the probability distribution expression of a composite function, we should respectively have an inverse function of Y=f(X) and Z=g(Y) in advance; further, there must be their explicit forms of inverse image.

Sometimes, it seems to be difficult for us to obtain these measurable transformations, or an explicit transformation is very complicated for treatment. I do not know whether there are some techniques to settle this matter.

Thanks.

Xunxue Cui : My pleasure. Yes, the superscript '-1' denotes inverse function. Further, if you know how to work with Wolfram Mathematica (or any other comparable program), then it is relatively easy to calculate inverse of functions -- through defining implicit functions. As for 'measurable transformations', my impression is that for your work Lebesque measure suffices, since you are working with real and possibly complex random variables. Therefore, the reference that I have given in my previous comment (by Klenke) should contains answers to all your questions.

Behnam Farid

I appreciate your detailed explanation.

According to your remark, I have read some contents in Klenke’s book (Probability Theory: A comprehensive Course (2nd edition)). On page 39 in this book, Theorem 1.101 describes the measure density expression of a mapping result when the density of a random variable is known.

This calculation expression is similar to your introduction in your first answer, except that the former uses a determinant operator “|det(…)|” in denominator, and is applied to the single one mapping. In my question, the variables and the mapping results are real-valued vector and matrix.

Your expression is right for double mappings in constructing a composite function, i.e.

φZ(z) = φX(h-1(z))/h'(h-1(z)) ≡ φX(g-1(f -1(z))/[h'(g-1(f -1(z))].

I do not know whether the absolute-valued determinant operator is necessary in denominator, regarding the real-valued random variable vector.

Thanks for providing very constructive comments and suggestions.

Xunxue Cui : You mention 'similar', but the correct wording would be 'identical', since the expression in Eq. (2) of my first comment on this paper is taken from exactly page 39 of the book by Klenke and cited as such.

As for the absolute value of the determinant to which you refer, it is necessary, because a probability distribution function must be non-negative [1]. This aspect is implicit in the expressions in Eqs (5a) and (5b) of my first comment on this page. Nonetheless, you have to go through my algebra and ensure that for your specific case this is also the case for the distribution function in Eq. (7), since I haven't been sufficiently meticulous on this issue [2] when writing my first comment on this page.

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[1] Note that when effecting transformations in evaluating multidimensional integrals, one always has the absolute value of the corresponding Jacobian determinant, an aspect tied to the non-negativity of the measure functions.

[2] A careful analysis requires looking into the existence of inverse functions. A function f(x) defined over an open interval is invertible if f '(x) ≠ 0 over that interval. When this is not the case, the inverse function is not unique, so that one will have to select the appropriate branch of a multivalued function in deducing the inverse function. In practice one avoids this aspect of selection by subdividing the interval under discussion into smaller intervals, the boundary separating the sub-intervals coinciding with the points where f '(x) = 0. In this way, by construction in each sub-interval f(x) is invertible. In my considerations, I neglected all the fine issues of this kind. Note that, in the context of measure theory, integrals are formally defined over positive functions so that in dealing with an arbitrary function, taking both positive and negative values, one has to first express this function as the difference of two positive functions; using linearity, the integral of the function of interest is then the difference of two integrals of positive functions.

Behnam Farid

Thank you for your detailed comments. Now I have understood the problem of the absolute determinant value. As Gaussian distribution is applicable for my issue, according to other references, the composite function is still distributed by the Gaussian probability distribution function, but with a different set of parameters in it from those in the original measurement equation.

In other words, if two Gaussian models are applied for the estimation result error of the two mappings, their inverse functions of two mappings are not necessary to derive. Of course, those parameters in the two new Gaussian models should be derived, which are determined by the error covariance of estimated variables based on the special problem.

It is worth mentioning that two linear transformations are adopted in my issue, after I have used a pseudo-linearization technique to construct the composite function. It is well known in the probability theory that the result of a linear transformation is still distributed by Gaussian density, if the error of variables in an equation is Gaussian.

Your suggestion on the deduction of inverse function is very important, particularly for the case that f’(x) equals zero at some points.

Your comments have given me great inspiration. Thanks.

Xunxue Cui : Thank you for the update, and, as regards my comments, my pleasure.